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Question Number 59872 by necx1 last updated on 15/May/19
$${solve}\:{the}\:{o}\:{d}\:{e} \\ $$$$\left(\mathrm{1}+{siny}\right){dx}=\left\{\mathrm{2}{y}\mathrm{cos}\:{y}−{x}\left({secy}+{tany}\right)\right\}{dy} \\ $$
Answered by ajfour last updated on 15/May/19
![(dx/dy)+x(sec y+tan y)=2ycos y e^(∫(( 1+sin y)/(cos y))dy) =e^(∫tan (y/2)dy) =e^(2ln sec (y/2)) = sec^2 (y/2) xsec^2 (y/2) = ∫((4ycos y)/((1+cos y)))dy xsec^2 (y/2) =4∫ydy−4∫((ydy)/(1+cos y)) xsec^2 (y/2) =4∫ydy−2∫ysec^2 (y/2)dy xsec^2 (y/2)=2y^2 −2[2ytan (y/2)−∫2tan (y/2)dy] xsec^2 (y/2)=2y^2 −4ytan (y/2)+8ln (sec (y/2))+c .](Q59874.png)
$$\frac{\mathrm{dx}}{\mathrm{dy}}+\mathrm{x}\left(\mathrm{sec}\:\mathrm{y}+\mathrm{tan}\:\mathrm{y}\right)=\mathrm{2ycos}\:\mathrm{y} \\ $$$$\mathrm{e}^{\int\frac{\:\mathrm{1}+\mathrm{sin}\:\mathrm{y}}{\mathrm{cos}\:\mathrm{y}}\mathrm{dy}} =\mathrm{e}^{\int\mathrm{tan}\:\frac{\mathrm{y}}{\mathrm{2}}\mathrm{dy}} =\mathrm{e}^{\mathrm{2ln}\:\mathrm{sec}\:\frac{\mathrm{y}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\:\mathrm{sec}\:^{\mathrm{2}} \frac{\mathrm{y}}{\mathrm{2}} \\ $$$$\mathrm{xsec}\:^{\mathrm{2}} \frac{\mathrm{y}}{\mathrm{2}}\:=\:\int\frac{\mathrm{4ycos}\:\mathrm{y}}{\left(\mathrm{1}+\mathrm{cos}\:\mathrm{y}\right)}\mathrm{dy} \\ $$$$\mathrm{xsec}\:^{\mathrm{2}} \frac{\mathrm{y}}{\mathrm{2}}\:=\mathrm{4}\int\mathrm{ydy}−\mathrm{4}\int\frac{\mathrm{ydy}}{\mathrm{1}+\mathrm{cos}\:\mathrm{y}} \\ $$$$\mathrm{xsec}\:^{\mathrm{2}} \frac{\mathrm{y}}{\mathrm{2}}\:=\mathrm{4}\int\mathrm{ydy}−\mathrm{2}\int\mathrm{ysec}\:^{\mathrm{2}} \frac{\mathrm{y}}{\mathrm{2}}\mathrm{dy} \\ $$$$\mathrm{xsec}\:^{\mathrm{2}} \frac{\mathrm{y}}{\mathrm{2}}=\mathrm{2y}^{\mathrm{2}} −\mathrm{2}\left[\mathrm{2ytan}\:\frac{\mathrm{y}}{\mathrm{2}}−\int\mathrm{2tan}\:\frac{\mathrm{y}}{\mathrm{2}}\mathrm{dy}\right] \\ $$$${x}\mathrm{sec}\:^{\mathrm{2}} \frac{{y}}{\mathrm{2}}=\mathrm{2}{y}^{\mathrm{2}} −\mathrm{4}{y}\mathrm{tan}\:\frac{{y}}{\mathrm{2}}+\mathrm{8ln}\:\left(\mathrm{sec}\:\frac{{y}}{\mathrm{2}}\right)+{c}\:. \\ $$$$ \\ $$
Commented by necx1 last updated on 18/May/19
$${yeah}.....\:{So}\:{grateful} \\ $$